Scenario: The owner of this network possesses a Class B license (190.123.0.0). E

Question

Scenario:

The owner of this network possesses a Class B license (190.123.0.0).

Each Network above is a populated network (each network has servers and workstations in it).

Routers 1, 2, 3 and 4 are non-compliant, meaning that they will not pass BootP and routing is to be statically configured. Networks A and B each have a non-WINS (Linux) machine that needs to have its name/IP address resolved on this Microsoft-based network.

Each network has one server and the following services are needed somewhere on the infrastructure: WINS (2 servers), DHCP (1 server) and DNS (2 servers)

All computers on each network need access to the Internet and the company maintains a web server on each network.

Do not worry about the Internet connection your ISP will provide that address.

Configure this network to meet the following requirements:

The Class B license must be subnetted to allow for the 4 subnets with a maximum number of machines on each subnet. Give the address of each subnet, the subnet mask to be used, the allowable machine addresses on each subnet and the broadcast address for each subnet.

List all the protocols that must be used and why each is used. No extra protocols are allowed since they would only waste bandwidth.

Configure nodes on each network (typically server machines) so that the necessary connectivity is achieved.

Whenever possible each machine should receive its IP address via DHCP and resolve its name to its IP address via DNS. Statically configure the IP addresses of those machines that require it. Assure that the non-WINS clients have their name resolved.

Fully configure the lone DHCP server to issue the IP addresses across the four subnets.

Deliverables

A subnet chart showing the Network ID, Gateway, Subnet Mask for each subnet

A routing table chart/table showing all static routes to the other networks

Router 1 Network A Network 190.123.0.0 Router 2 Router 3 Network B Network C Router 4 To the Internet Network D

Explanation / Answer

Following lines provide you the required solution for the first question.

In order to divide given class B network in 4 subnets, we need to extend the no. of subnet bits by 2. Now, number of network bits are increased to 18 bits and no. of hosts bits are 14. So, subnet mask of each subnet will be 255.255.192.0 (i.e., 11111111 11111111 11000000 00000000).

So, for network A, no. of hosts will be (2^14 - 2). Network address will be 190.123.0.0 (190.123. 00000000. 00000000) and broadcast address will be (190.123.0011 1111. 1111 1111) i.e., 190.123. 63. 255 and allowable machine address for this network is 190.123.0.1 to 190.123.63.254.

Similarly, for network B, no. of hosts will be (2^14 - 2). Network address will be 190.123.64.0 (190.123. 01000000. 00000000) and broadcast address will be (190.123.0111 1111. 1111 1111) i.e., 190.123. 127. 255 and allowable machine address for this network is 190.123.64.1 to 190.123.127.254.

For network C, no. of hosts will be (2^14 - 2). Network address will be 190.123.128.0 (190.123. 10000000. 00000000) and broadcast address will be (190.123.011 1111. 1111 1111) i.e., 190.123. 191. 255 and allowable machine address for this network is 190.123.128.1 to 190.123.191.254.

Finally, for network D, no. of hosts will be (2^14 - 2). Network address will be 190.123.192.0 (190.123. 11000000. 00000000) and broadcast address will be (190.123.1111 1111. 1111 1111) i.e., 190.123. 255. 255 and allowable machine address for this network is 190.123.192.1 to 190.123.255.254.

The following table summarize above work:

Network Network ID Network Range Broadcast Address Network A 190.123.0.0 190.123.0.1 - 190.123.63.254 190.123.63.255 Network B 190.123.64.0 190.123.64.1 - 190.123.127.254 190.123.127.255 Network C 190.123.128.0 190.123.128.1 - 190.123.191.254 190.123.191.255 Network D 190.123.192.0 190.123.192.1 - 190.123.255.254 190.123.255.255

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